MEI C2 Study Resources Core2 Logarithms And Exponentials 1-2

MEI Core 2 Logarithms and Exponentials Section 1: Introduction to logarithms Study Plan Background Before the introduction of calculators, students used to carry log tables around with them. They were invaluable aids when long multiplication, long division sums had to be done, in the same way that calculators are very useful now. However, logarithms have a much wider use than in making arithmetic easier to carry out. They crop up in many areas of life, especially in the sciences.

Detailed work plan 1. Before you start reading this chapter, go back to chapter 5 and revise the work on indices. This will help you follow the work in this chapter. 2. Read the introduction to logarithms on pages 319 - 320. You will need to be able to switch quickly from equations in log form to ones in index form. There is a worked example on this in the Notes and Examples. 3. You can see some further examples using the Flash resource Basic logarithms. 4. Before reading any further, try questions 1, 2 and 3 of Exercise 11A. This will ensure that you have grasped the relationships between index form and logarithms. 5. For additional practice, try the interactive questions Bases of logs. 6. Read pages 321 to 324 especially example 11.2. This is a question that you could not have tackled using only the work in chapter 5. On page 321 you can see one way in which the laws of logarithms are derived from the laws of indices; there is a different method shown in the Notes and Examples, together with some further examples there. 7. Try the Flash resources Laws of logarithms, The graph of y = kax and Solving ax = b. You may also find the Mathcentre video Logarithms useful. 8. Exercise 11A Try questions 4, 5, 6*, 7*, 8*, 9, 10. 9. For some additional practice in using the laws of logarithms, try the Logarithms puzzle. 10. For extra practice try the interactive questions Solving for powers.

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MEI Core 2 Logarithms and exponentials Section 1: Introduction to logarithms Notes and Examples These notes contain subsections on  Indices and logarithms  The laws of logarithms  Exponential functions  Solving exponential equations using logarithms  An old practical application of logarithms (extension work)

Indices and logarithms The important thing to remember about logarithms is that, although they appear to be a new topic, they are simply about writing what you already know about indices in a different way. If you find it difficult to work out the meaning of a statement involving logarithms, it can be simpler to change the statement into the equivalent statement involving indices.

log a b  x  a x  b . To remember this, notice that a is both the base of the logarithm and the base of the index, and x, the logarithm, is the index. The value of log a b is the answer to the question: “What power must I raise a to in order to get b?”

Example 1 (i) Find log 4 2 (ii) Find x, where log5 x   12 Solution (i) The statement log 4 2  x is equivalent to 4x  2 . 1

Since 4 2  2 , then x must be So log 4 2  12 .

1 . 2

(ii) The statement log5 x   12 is equivalent to 5 1 So x  . 5



1 2

 x.

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MEI C2 Logs Section 1 Notes and Examples You can also look at the Flash resource Basic logarithms. For practice in understanding statements involving logarithms, try the interactive questions Bases of logarithms.

The laws of logarithms The laws of logarithms are:

log x  log y  log xy x log x  log y  log y log x n  n log x

These can be proved using the laws of indices:

It’s a worthwhile exercise to try to work through these proofs

First convert into index notation: log c x  a  c a  x

log c y  b  cb  y To prove the first law: a b

c c  xy  c  xy  log c xy  a  b a b

Using the laws of indices

 log c xy  log c x  log c y Similarly for the second law: x ca x   c a b  b y c y x  log c  a  b y x  log c  log c x  log c y y For the third law:

log c x n  a  c a  x n  ca n  x a n  n log c x  a  log c x 

 n log c x  log c x n As the first two laws of indices require the indices to have the same base, then the first two laws of logarithms require the logarithms to have the same base.

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MEI C2 Logs Section 1 Notes and Examples Example 2

x3 y (i) Write log in terms of log x, log y and log z. z (ii) Write 2log a  log b  13 log c as a single logarithm. Solution x3 y (i) log  log x3  log y  log z z  log x3  log y  log z1 2

 3log x  log y  12 log z (ii) 2 log a  log b  13 log c  log a 2  log b  log c1 3

 log a 2  (log b  log 3 c )  log

a2 b3 c

You can also look at the Flash resource Laws of logarithms, and you may find the Mathcentre video Logarithms useful.

For some practice in using the laws of logarithms, try the Logarithms puzzle.

Exponential functions An exponential function is any function of the form y  a x . The graphs below show some different exponential functions. x x y  5x y  2 y  1.5

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MEI C2 Logs Section 1 Notes and Examples Exponential functions are the inverse of logarithm functions: the function y  a x is the inverse of the function y  log a x . This relationship is very useful for solving equations, as shown below.

You can explore graphs of this type using the Flash resource The graph of y = kax.

Solving exponential equations using logarithms Logarithms are very useful for solving equations where the unknown variable is an index, such as the equation 2 x  10 . Many equations are solved using inverse functions, for example you solve the equation x + 3 = 5, in which addition is applied to the unknown variable, by subtracting 3 from each side. Similarly you solve the equation x² = 10 by using the square root function, which is the inverse of the square function. An equation like 2 x  10 involves an exponential function of x. So to solve this equation, it follows that you need to use the inverse of the exponential function, which is the logarithm function. This is shown in the next example.

Example 3 Solve the following equations. (i) 2 x  10 (ii) 32 x1  4 (iii) 0.21 x  2 Solution (i) 2 x  10

log 2 x  log10 x log 2  log10 log10 x  3.32 log 2 (ii) 32 x 1  4

log 32 x 1  log 4 (2 x  1) log 3  log 4 log 4 2x 1  log 3 1  log 4  x   1  1.13 2  log 3 

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MEI C2 Logs Section 1 Notes and Examples (iii) 0.21 x  2

log 0.21 x  log 2 (1  x) log 0.2  log 2 log 2 1 x  log 0.2 log 2 x  1  1.43 log 0.2

You can see similar examples using the Flash resource Solving a^x = b. For practice in this type of problem, try the interactive questions Logarithms: Solving powers.

An old practical application of logarithms Before calculators existed, logarithms were used to make calculations easier. For example, suppose you had to divide 1432627 by 967253. You could do this by long division, but it would take a long time and the chances of making a mistake would be quite high. So you would apply the second law of logarithms: log (1432627  967253) = log 1432627  log 967253 To do the calculation, you would have to find the log to base 10 of the two numbers, subtract the results, and then find the inverse log of the answer. You would have to find the values of log 1432627 and log 967253 from a book of tables. Unfortunately most tables would only tell you the values of log x for values of x between 10 and 99. So you would then use the fact that

log 1432627  log(14.32627 100000)  log14.32627  log100000  log14.32627  5 You would then use the tables to find the value of log 14.33 (which is as accurate as most tables would give you). This would give a value for log 1432627 of 6.1562. You would then go through a similar process to find log 967253.

log 967253  log 96.7253  log10000  1.9855  4  5.9855 Next you would subtract these two logarithms (without a calculator of course!), giving 0.1707.

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MEI C2 Logs Section 1 Notes and Examples Now you would have to find the number whose logarithm is 0.1707. Inverse log tables usually give values between 1 and 10. 0.1707  1.1707  1  log 14.81  log 10

14.81 10  log1.481  log

So 1432627  967253 = 1.481. Most pupils did not understand the theory behind these calculations; they just followed a set of instructions to use the tables of logarithms and work out the calculation. Even after calculators became widely available, it was several years before this technique was removed from examination syllabuses! Logarithms were also the basis of slide rules, which were also used before calculators existed to work out calculations quickly.

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Core 2 Logarithms and exponentials Section 1: Introduction to logarithms Crucial points 1. Make sure that you know the equivalent log relationships It can be difficult to develop a “feel” for logarithms. Keep the equivalent relationships log a b = c ⇔ a c = b firmly in mind, remembering that the base of the logarithm is also the base of the index. The value of log a b is the answer to the question: “What power must I raise a to in order to get b?” 2. Remember the log laws Make sure that you know the laws of logarithms as this can save time in exams because you do not need to look in the formulae book.

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MEI Core 2 Logarithms and exponentials Section 1: Introduction to logarithms Exercise 1.

Rewrite each of these statements as a logarithm. (i) 103 = 1000 (ii) 27 = 128 1 1 (iii) 10 3 = 3 10 2−3 = (iv) 8 1 3 − 1 5 2= (vi) 3 2 = 27 (v) 5

2.

Find the values of the following: log 2 16 (i)

(ii)

(iii)

log 6 1

(iv)

(v)

log 5 5

(vi)

(vii)

log8 4

(viii)

3.

log10 1000000 ⎛1⎞ log 4 ⎜ ⎟ ⎝4⎠ ⎛ 1 ⎞ log 3 ⎜ ⎟ ⎝ 27 ⎠ ⎛ 1 ⎞ log 2 ⎜ ⎟ ⎝ 32 ⎠

Find the value of x in each of the following: (i) log 2 x = −5 (ii) log 3 x = (iii)

log x 64 = 2

(iv)

3 2

⎛ 1 ⎞ 1 log x ⎜ ⎟= ⎝ 5⎠ 2

4.

Write the following as a single logarithm: (i) log 2 + log 3 (ii) log 10 – log 2 (iii) 3 log 5 (iv) 2 log 3 – 4 log 2 1 1 (v) log 3 − log 4 (vi) 2 log a + 5 log b – 3 log c 2 4

5.

Write the following in terms of log 2 and log 3:

6.

(i)

log 12

(ii)

(iii)

log 54

(iv)

Solve the following equations: (i) 2 x = 18 (iii) 1.5 x = 0.001

(ii) (iv)

⎛ 16 ⎞ log ⎜ ⎟ ⎝ 27 ⎠ 3 log 16

5x = 100 10 x = 2

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MEI Core 2 Logarithms and exponentials Section 1: Introduction to logarithms Solutions to Exercise 1. (i)

103  1000  log10 1000  3

(ii) 2 7  128  log2 128  7 1

(iii) 10 3  3 10 (iv) 2 3 

1 8

(v) 5  2  1

 log2 1 5

3

(vi) 3 2  27

2. (i)

 log10 3 10  1 8

1 3

 3

 log5

1   21 5

 log3 27 

3 2

x  log2 16  2 x  16  x  4 so log2 16  4

(ii) x  log10 1000000  10 x  1000000  x  6 so log10 1000000  6 (iii) x  log6 1  6 x  1  x  0 so log6 1  0

1 (iv) x  log4    4 x   4 1 so log4    1 4 (v)

x  log5 5 so log5

5 

 x  1

1 4

5x  5

x

1 2

1 2

 1   3x  (vi) x  log3    27   1  so log3    3  27 

1 27

 x  3

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MEI C2 Logarithms Section 1 Exercise solutions (vii) x  log8 4  8 x  4  2 3 x  2 2 so log8 4  23

 1  x (viii) x  log2   2   32   1  5 so log2   2  32 

3. (i)

 2 5  x

log2 x  5

(ii) log3 x 

3 2

3

 32  x

1  32

x

x

1 2

2 3

 2 2 5

5

 x   52

1 32

 x  27

(iii) logx 64  2  x 2  64  x  8

 1  (iv) log x    5

4. (i)

1 2

1

 x2 

1 5

x

1 5

log2  log3  log(2  3)  log6

(ii) log 10  log2  log

10  log 5 2

(iii) 3log5  log5 3  log125 (iv) 2 log3  4log2  log32  log2 4  log

(v)

1 2

1

1

log3  41 log4  log3 2  log4 4  log

32 9  log 16 4 2

3  log 2

3 2

(vi) 2 log a  5 logb  3log c  log a 2  logb 5  log c 3  log

5. (i)

a 2b 5 c3

log12  log(2 2  3)  log2 2  log3  2 log2  log3

 16   24  4 3  log (ii) log    33   log2  log3  4log2  3log3 27     (iii) log 54  log  2  33  2  log2 2  log3 2  21 log2  23 log3 1

1

3

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MEI C2 Logarithms Section 1 Exercise solutions 1

1 3 32 (iv) log  log 4  log3 2  log2 4  21 log3  4log2 16 2

6. (i)

2 x  18 log 2 x  log 18

x log 2  log 18 x

log 18  4.17 (3 s.f.) log 2

(ii) 5 x  100

log 5 x  log 100

x log 5  log 100 x

log 100  2.86 (3 s.f.) log 5

(iii) 1.5 x  0.001

log 1.5 x  log0.001

x log 1.5  log0.001 x

log0.001  17.0 (3 s.f.) log 1.5

(iv) 10 x  2

log 10 x  log 2

x log 10  log 2 x

log 2  0.301 (3 s.f.) log 10

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Logarithms Hexagonal Jigsaw The three diagrams on the following pages will ultimately fit together to form a large hexagon. Before you start, the three diagrams must be cut along the lines to make twenty-four equilateral triangles. For the triangles to be fitted together, you must find two expressions that are equivalent to one another. To build up the puzzle, place the edges on which equivalent expressions are written together, so that the triangles are joined along this edge. Begin by just finding pairs of matching expressions and placing them edge-to-edge. As you progress you will find that all of the pieces will eventually link up to form a large hexagon. By all means try this by yourself, but it is really designed to be a group activity. Working with other students will help to highlight any misconceptions you may have, as well as making the task more enjoyable. At first glance this may appear to be an easy undertaking, but you will find it takes quite a lot of thought and errors can be made very easily. Make sure you check each coupling, or you could be left with one or two pieces that appear to fit nowhere.

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og a

l

6

log a

2+

log a

3

log a 7 8

x

5

7

log a

7+

log a 12

log a

a

log a 2 − log a 7

−l og

y x a

7

og a

log a log

x

3

l + 5

a

x

log a 9

x−

1 log a 9 2

ga

log a

y

log a 5 − log a 20

g o l 4

2 lo

a

a

3 log a 2

log

−3 log

3

log a 18

6

log a

12

ga o l −

4

y+

log

a

y

3

log a

x

− lo

x2 log a y

log

x

5

ga

x

3

og a

42

a

a

7 log a 6

log

2l

log a 4

0

x

log a

12

a

8 l o g

7

2 log a 3

log a

a

og a

3

8

8

l

− 3 log a 2

4−

log a 3

a

2 lo g

30

log

1 3

2+

3−

log a

3 lo ga

a

log a

log a

32

log

log x a y + log

a

x

y

36

x3

log a

log

72

a

a

1 3 log

8

5 log a 2

2 log a 7

log

log a

0.2

a

(x

x

g o l 2

+ x a

2

ga o l 5

y

1 log a 2

7

2

15

1 2 loga

a

6log a x

log

4

log a x 2 y 5

log a

5

4

log a

6

2

7 0.3

log a

log a (x − 1)

a

ga + lo

ga o l −

log a

3+

log

lo g a0 . 2 5

1

log a

5+

1 8

a

log a

log

−1

)−

log

a

(x +

1)

MEI Core 2 Logarithms and exponentials Section 1: Introduction to logarithms Multiple Choice Test 1) The value of log2 32 is: (a) 5 (c) 16 (e) I don’t know

(b) 6 (d) 7

⎛ 1 ⎞ 2) The value of log 3 ⎜ ⎟ is: ⎝ 3⎠ (a) –1 (c) –2 (e) I don’t know

(b) 12 (d) − 12

3) The statement a b = c is equivalent to the statement: (a) c = log a b (c) a = log b c (e) I don’t know

(b) b = log a c (d) c = log b a

4) If log 3 x = −3 then x = (a) 27 (c)

3

3

1 27 1 (d) 3 3 (b)

(e) I don’t know

5)

a log b + b log a =

(a) log(a bb a ) (c) (a + b) log(ab) (e) I don’t know

(b) ab log(ab) (d) log(a b + b a )

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MEI C2 Logarithms Section 1 MC test 6)

log x − 2 log y + 12 log z =

⎛ xz ⎞ (a) log ⎜ ⎟ ⎝ 4y ⎠

⎛x z (b) log ⎜⎜ 2 ⎝ y

⎛ xz ⎞ (c) log ⎜ ⎟ ⎝ y⎠ (e) I don’t know

(d) log x − y 2 + z

(

⎞ ⎟⎟ ⎠

)

7) If a x = b , the value of x is given by

⎛b⎞ (a) x = log ⎜ ⎟ ⎝a⎠ b (c) x = log a (e) I don’t know

(b) x = log b − log a (d) x =

log b log a

8) If p −2 x = q , the value of x is given by log q 2 log p ⎛q⎞ 1 (c) x = − log ⎜ ⎟ 2 ⎝ p⎠ (e) I don’t know

(a) x = −

log q +2 log p ⎛q⎞ (d) x = log ⎜ ⎟ + 2 ⎝ p⎠ (b) x =

9) Which of the following statements are true? (i) log 2 8 = 3 (ii) log 2 8 = −3 (iv) log8 2 = 3 (v) log8 2 = −3 (a) (i) and (v) (c) (i) and (vi) (e) I don’t know

(iii) (vi)

log 2 8 = 13 log8 2 = 13

(b) (iii) and (vi) (d) (ii) and (vi)

10) A geometric series has common ratio 0.9. The smallest number of terms required for the sum of the series to be greater that 99.9% of the sum to infinity is (a) 44 (c) 43 (e) I don’t know

(b) 65 (d) 66

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MEI Core 2 Logarithms and Exponentials Section 2: Modelling curves using logarithms Study Plan Background Modelling in mathematics underpins much that you do and this section is particularly relevant when seeking relationships between variables. As you have seen with earlier work, dealing with a straight line and gaining information from it is easier than dealing with curves. This section shows how logs can be used to reduce a relationship to straight line form and hence make it easier to plot and to evaluate.

Detailed work plan 1. Read examples 11.4, 11.5 and 11.6 on pages 326 – 331. There are two further examples in the Notes and Examples. 2. For practice in these, try the interactive questions Equations from log-log graphs and Equations from log-x graphs. 3. Try the questions listed below. Structure your answers carefully so that you can see clearly the quantities that you need to plot and their relationship to the gradient and the intercept of the line. Exercise 11B 2, 3*, 4*, 6, 8*, 9, 11, 12*

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MEI Core 2 Logarithms and exponentials Section 2: Modelling curves using logarithms Notes and Examples These notes contain subsections on: n  Modelling curves of the form y = kx x  Modelling curves of the form y = ka

Modelling curves of the form y  kx n When you collect data from an experiment, you may want to find a relationship between two variables, such as the speed of a moving object at a particular time, or temperature of an object and its distance from a heat source. You may plot a graph of one variable against another to help find this relationship. However, unless the graph is a straight line, it may be difficult to see the relationship from the graph. The graphs of y  x 2 , y  x3 etc. look quite similar for x  0, and as the experimental data may not be very accurate, it can be impossible to tell with any certainty what would be the best graph to model the data. This is where logarithms can be very useful. If the relationship is of the form y  kx n , then plotting log y against log x gives a straight line graph.

y  kx n  log y  log kx n

This is of the form y = mx + c, so plotting log y against log x gives a straight line with gradient n and intercept log k.

 log y  log k  log x  log y  log k  n log x  log y  n log x  log k n

The value of n is therefore the gradient of the graph, and the value of k is found by taking the intercept of the graph and finding its inverse logarithm (i.e. 10intercept if you are using logs to base 10).

Example 1 The relationship between two variables x and y is believed to be of the form y  kx n , where k and n are constants. In an experiment, the following values of x and y are recorded. x y

1 1.98

2 1.39

3 1.16

4 1.01

5 0.91

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6 0.82

7 0.75

8 0.72

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MEI C2 Logarithms Section 2 Notes and Examples Verify that the model y  kx n is appropriate and find the approximate values of the constants k and n. Solution Taking logarithms:

y  kx n log y  log kx n

log y  log k  log x n log y  log k  n log x If log y is plotted against log x, this is the equation of a straight line graph with gradient n and intercept log k. Plot the values of log y against log x: x log x y log y

1 0 1.98 0.30

2 0.30 1.39 0.14

3 0.48 1.16 0.06

4 0.60 1.02 0.01

5 0.70 0.91 -0.04

6 0.78 0.82 -0.09

7 0.85 0.75 -0.12

8 0.90 0.72 -0.14

0.18 0.36

Since the graph is approximately a straight line, the relationship y  kx n is an appropriate model. 0.18  0.5 Gradient = n =  0.36 Intercept  log k  0.3  k  100.3  2

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MEI C2 Logarithms Section 2 Notes and Examples The relationship is approximately y  2 x 0.5 

2 x

For practice in problems similar to this, try the interactive questions Equations from log-log graphs.

Modelling curves of the form y  ka x Similarly, if the relationship is of the exponential form y  ka x , then plotting log y against x gives a straight line graph.

y  ka x  log y  log ka x

This is of the form y = mx + c, so plotting log y against x gives a straight line with gradient log a and intercept log k.

 log y  log k  log a  log y  log k  x log a x

 log y   log a  x  log k

The value of a is found by taking the gradient of the graph and finding its inverse logarithm (i.e. 10gradient if you are using logs to base 10), and the value of k is found by taking the intercept of the graph and finding its inverse logarithm (i.e. 10intercept if you are using logs to base 10).

Example 2 The relationship between two variables p and q is believed to be of the form q  ab p , where p and q are constants. In an experiment, the following values of p and q are recorded. p q

1.5 12

2.0 19

2.5 30

3.0 46

3.5 74

4.0 116

Verify that the model q  ab p is appropriate, and estimate the values of a and b. Solution Taking logarithms:

q  ab p log q  log ab p

log q  log a  log b p log q  log a  p log b If log q is plotted against p, this is the equation of a straight line with gradient log b and intercept log a.

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MEI C2 Logarithms Section 2 Notes and Examples Plot the values of log q against p: p q log q

1.5 12 1.08

2.0 19 1.28

2.5 30 1.48

3.0 46 1.66

3.5 74 1.87

4.0 116 2.06

1.25

3.2

Since the graph is approximately a straight line, the relationship q  ab p is an appropriate model.

1.25  b  101.25 / 3.2  2.5 3.2 Intercept = log a = 0.5  a  100.5  3.2 Gradient = log b =

The relationship is approximately q  3.2  2.5 p

You do not need to remember the details of what to plot and what to do with the gradient and intercept of the graph – all you need to do is to take logs of both sides of the suggested relationship and apply the laws of logarithms to obtain a relationship of the form y = mx + c, as shown above for each of the relationships y  kx n and y  ka x . Once you have done this, you can see what you need to plot and how to find the values of the constants.

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MEI C2 Logarithms Section 2 Notes and Examples If the graph is not a straight line, then the suggested model is not an appropriate one (or perhaps the experimental results are not sufficiently accurate). For practice in problems like the one above, try the interactive questions Equations from log-x graphs.

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Core 2 Logarithms and exponentials Section 2: Modelling curves using logarithms Crucial points 1. You should be able to convert polynomial functions into linear equations using logs Make sure that you can show how the relationships y  kx n and

y  ka x can be written in the form y = mx + c by using logarithms. (See the “Notes and Examples”). 2. Make sure you plot the right quantities against each other Check the linear form of the relationship to see what you need to plot Plot log y against log x y  kxn  log y  log k  n log x

y  ka x  log y  log k  x log a

Plot log y against x

3. Remember to use inverse logs where appropriate to find the unknown constant Make sure you know the relationships between the unknown constants and the gradient and intercept of the graph that you have plotted. y  kxn  log y  log k  n log x Gradient = n, intercept = log k

y  ka x  log y  log k  x log a Gradient = log a, intercept = log k

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MEI Core 2 Logarithms and exponentials Section 2: Modelling curves using logarithms Exercise 1. Two variables s and t are related by the formula s = at c , where a and c are constants. (i) Show that this relationship can be written as log s = log a + c log t . (ii) Explain why the model can be tested by plotting log s against log t. Values of s and t are recorded in an experiment. 9 5

s t

13 10

16 15

18 20

20 25

22 30

(iii) Plot the graph of log s against log t and use your graph to estimate the values of a and c. 2. Two variables a and b are related by the formula b = mn a , where m and n are constants. (i) Show that this relationship can be written as log b = log m + a log n . (ii) Explain why the model can be tested by plotting log b against a.

In an experiment, the following values of a and b are obtained. 0.5 4.5

a b

1.0 4.0

1.5 3.6

2.0 3.2

2.5 2.9

3.0 2.6

3.5 2.3

(iii) Plot the graph of log b against a and use your graph to estimate the values of m and n. 3. The relationship between two variables x and y is believed to be of the form y = kx n , where k and n are constants. In an experiment, the following values of x and y are recorded. x y

1 3

2 8

3 16

4 24

5 34

6 44

7 56

Plot the graph of log y against log x and explain why this tells you that the model y = kx n is appropriate. (ii) Use your graph to estimate the values of k and n. (iii) Estimate the value of y when x = 10. (i)

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MEI C2 Logs and exps Section 2 Exercise 4. The relationship between two variables p and q is believed to be of the form q = ab p , where a and b are constants. In an experiment, the following values of p and q are recorded. p q

1.2 2.5

3.4 3.7

5.7 5.8

6.2 6.1

7.4 7.7

9.8 11.9

Plot the graph of log q against p, and explain why this tells you that the model q = ab p is appropriate. (ii) Use your graph to estimate the values of a and b. (iii) Estimate the value of q when p = 12. (i)

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MEI Core 2 Logarithms and exponentials Section 2: Modelling curves using logarithms Exercise 1. Two variables s and t are related by the formula s = at c , where a and c are constants. (i) Show that this relationship can be written as log s = log a + c log t . (ii) Explain why the model can be tested by plotting log s against log t. Values of s and t are recorded in an experiment. 9 5

s t

13 10

16 15

18 20

20 25

22 30

(iii) Plot the graph of log s against log t and use your graph to estimate the values of a and c. 2. Two variables a and b are related by the formula b = mn a , where m and n are constants. (i) Show that this relationship can be written as log b = log m + a log n . (ii) Explain why the model can be tested by plotting log b against a.

In an experiment, the following values of a and b are obtained. 0.5 4.5

a b

1.0 4.0

1.5 3.6

2.0 3.2

2.5 2.9

3.0 2.6

3.5 2.3

(iii) Plot the graph of log b against a and use your graph to estimate the values of m and n. 3. The relationship between two variables x and y is believed to be of the form y = kx n , where k and n are constants. In an experiment, the following values of x and y are recorded. x y

1 3

2 8

3 16

4 24

5 34

6 44

7 56

Plot the graph of log y against log x and explain why this tells you that the model y = kx n is appropriate. (ii) Use your graph to estimate the values of k and n. (iii) Estimate the value of y when x = 10. (i)

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MEI C2 Logs and exps Section 2 Exercise solutions 4. The relationship between two variables p and q is believed to be of the form q = ab p , where a and b are constants. In an experiment, the following values of p and q are recorded. p q

1.2 2.5

3.4 3.7

5.7 5.8

6.2 6.1

7.4 7.7

9.8 11.9

Plot the graph of log q against p, and explain why this tells you that the model q = ab p is appropriate. (ii) Use your graph to estimate the values of a and b. (iii) Estimate the value of q when p = 12. (i)

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MEI C2 Logs and exps Section 2 Exercise solutions Solutions to Exercise 1. (i)

s = at c

Taking logarithms of both sides: log s = log(at c ) = log a + logt c = log a + c logt

(ii) Since log a and c are constants, the equation log s = log a + c logt is the equation of a straight line, in which the variables are log t and log s, and which has gradient c and intercept log a. So if the model is appropriate, plotting log s against log t will give an approximate straight line. (iii)

s t log s log t 2

9 5 0.95 0.70

13 10 1.11 1

16 15 1.20 1.18

18 20 1.26 1.30

20 25 1.30 1.40

22 30 1.34 1.48

log s

1 .5

1

0.4 0.8

0.5

log t 0.5

1

1 .5

2

Equation of graph is log s = log a + c logt 0.4 = 0.5 , so c = 0.5 Gradient = 0.8 Intercept = 0.6, so log a = 0.6 ⇒ a = 10 0.6 ≈ 4 .

2. (i)

b = mn a logb = log(mn a ) = log m + log n a = log m + a log n

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MEI C2 Logs and exps Section 2 Exercise solutions (ii) The equation logb = log m + a log n is the equation of a straight line, in which the variables are log b and a, and which has gradient log n and intercept log m. So if the model is appropriate, then plotting log b against a will give an approximate straight line graph. (iii)

a b

0.5 4.5 0.62

log b 1

1.0 4.0 0.60

1.5 3.6 0.56

2.0 3.2 0.51

2.5 2.9 0.46

3.0 2.6 0.41

3.5 2.3 0.36

log b

0.9 0.8 0.7 0.6 0.5

0.18

0.4

2.0

0.3 0.2 0.1

a 0.5

1

1 .5

2

2 .5

3

3 .5

4

Equation of graph is logb = log m + a log n 0.18 Gradient = − = −0.09 , so log n = −0.09 ⇒ n = 10 −0.09 ≈ 0.8 2 Intercept = 0.68 , so log m = 0.68 ⇒ m = 10.68 ≈ 4.8

3. (i)

x y log x log y

1 3 0 0.48

2 8 0.30 0.90

3 16 0.48 1.20

4 24 0.60 1.38

© MEI, 12/08/08

5 34 0.70 1.53

6 44 0.78 1.64

7 56 0.85 1.75

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MEI C2 Logs and exps Section 2 Exercise solutions 2

log y

1 .5

1

0.6 0.4

0.5

log x 0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

y = kx log y = log k + n log x This is the equation of a straight line, with variables log y and log x. Since the points form an approximate straight line, the model is appropriate. n

(ii) log y = log k + n log x is the equation of a straight line with gradient n and intercept log k. 0.6 = 1.5 ⇒ n = 1.5 Gradient = 0.4 Intercept = 0.47 so log k = 0.47 ⇒ k = 10 0.47 ≈ 3 (iii) y = 3 x 1.5 When x = 10, y = 3 × 10 1.5 ≈ 95

4. (i)

p q log q

1.2 2.5 0.40

3.4 3.7 0.57

5.7 5.8 0.76

6.2 6.1 0.79

© MEI, 12/08/08

7.4 7.7 0.89

9.8 11.9 1.08

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MEI C2 Logs and exps Section 2 Exercise solutions

1 .2

log q

1

0.4

0.8 0.6

5

0.4 0.2 p 1

2

3

4

5

6

7

8

9

10

q = ab p log q = log a + p logb This is the equation of a straight line, with variables log q and p. Since the points form an approximate straight line, the model is appropriate. (ii) log q = log a + p logb is the equation of a straight line with gradient log b and intercept log a. 0.4 Gradient = = 0.08 , so logb = 0.08 ⇒ b = 10 0.08 ≈ 1.2 5 Intercept = 0.3 , so log a = 0.3 ⇒ a = 10 0.3 ≈ 2 (iii) q = 2 × 1.2 p When p = 12, q = 2 × 1.2 12 ≈ 17.8

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MEI Core 2 Logarithms and exponentials Section 2: Modelling curves using logarithms Multiple Choice Test Questions 1 and 2 are about an experiment in which the relationship between two variables N and t is believed to be of the form N = abt . A student plots an appropriate graph and finds that this gives a straight line with gradient m and intercept c.

1) The graph which the student plotted is (a) N against t (c) N against log t (e) I don’t know

(b) log N against log t (d) log N against t

2) The values of a and b are given by (a) a = 10c , b = 10m (c) a = m, b = 10c (e) I don’t know

(b) a = 10m , b = 10c (d) a = 10c , b = m

Questions 3 and 4 are about an experiment in which the relationship between two variables y and x is believed to be of the form y = kx n . A student plots an appropriate graph and finds that this gives a straight line with gradient m and intercept c.

3) The graph which the student plots is (a) y against log x (c) log y against log x (e) I don’t know

(b) log y against x (d) y against x

4) The values of n and k are given by (a) n = 10m , k = 10c (c) n = 10c , k = 10m (e) I don’t know

(b) n = 10c , k = m (d) n = m, k = 10c

Questions 5 and 6 are about the variables p and q which are believed to be connected by a relationship of the form p = Aq n . Data is collected and an appropriate graph plotted, which is a straight line with gradient -2 and intercept 0.5.

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MEI C2 Logs and exps Section 2 MC test 5) The value of A is approximately (a) -2 (c) 3.2 (e) I don’t know

(b) 0.5 (d) 0.01

6) The value of n is approximately (a) -2 (c) 0.01 (e) I don’t know

(b) 3.2 (d) 0.5

Questions 7 and 8 are about the variables s and t which are believed to be connected by a relationship of the form s = ka t . Data is collected and an appropriate graph plotted, which is a straight line with gradient 0.8 and intercept 0.3.

7) The value of k is approximately (a) 0.8 (c) 2.0 (e) I don’t know

(b) 6.3 (d) 0.3

8) The value of a is approximately (a) 0.8 (c) 0.3 (e) I don’t know

(b) 2.0 (d) 6.3

9) In an experiment, data is collected for two variables x and y. The graph of log y is plotted against log x and is found to be approximately a straight line with gradient 2 and intercept 0.6. The relationship between x and y is approximately given by (a) y = 4 x 2 (c) y = 0.6 x 2 (e) I don’t know

(b) y = 4 × 2 x (d) y = 0.6 × 2 x

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MEI C2 Logs and exps Section 2 MC test 10) In another experiment, data is collected for two variables p and q. The graph of log q is plotted against p and is found to be approximately a straight line with gradient 0.3 and intercept –1. The relationship between p and q is approximately given by (a) q = 10 × 2 p

(b) q = 10 p 2

2p 10 (e) I don’t know

(d) q =

(c) q =

p2 10

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MEI Core 2 Logarithms and Exponentials Chapter Assessment 1. Find the value of: (i) log10 1000000

(ii)

(iv) log 4 32

(v)

2. Write as a single logarithm: (i) 2 log a + 3log b (ii)

⎛1⎞ log 3 ⎜ ⎟ ⎝ 81 ⎠ log a a

(iii)

log 2 2

(vi)

log a 1

log x − 3log y + 4 log z

[6]

3. Express the following in terms of log p, log q and log r. p pq (i) log log 2 (ii) r r 4. Solve the following equations; (i) 2 x = 7 (ii) 5 x = 10

(iii)

[6]

[6]

32 x = 5

[9]

5. Alice puts £500 in a savings account, at a fixed interest rate of 5% per year, when her grandson Harry is born. Interest is added to the account on Harry’s birthday each year. The amount, P, in the account after n years is given by:

P = 500 ×1.05n How old will Harry be when the amount in the savings account first exceeds £1000? [6] 6. In an experiment, the number of bacteria, N, in a culture was estimated at time t days after the measurements started. The results were as follows: t N

1 120

2 170

3 250

4 400

5 620

6 910

It is believed that the relationship between N and t can be expressed in the form N = abt where a and b are constants. (i) (ii) (iii) (iv)

Explain how this can be tested by plotting log N against t. Make out a table of values of log N and draw the graph. Use your graph to estimate the values of a and b. Estimate the number of bacteria present after 20 days.

© MEI, 13/08/08

[2] [3] [3] [1]

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MEI C2 Logs and exponentials Assessment 7. It is believed that two quantities, x and y, are connected by a relationship of the form y = kx n , where k and n are constants. In an experiment, the following data were produced. x y (i) (ii) (iii)

5 9

10 24

15 48

20 69

25 102

30 131

35 166

Explain how the form of the relationship can be tested by plotting log y against log x. [2] Make out a table of values of log x and log y and plot the graph. [3] Use your graph to estimate the values of k and n. [3]

Total 50 marks

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